Prove that if $G$ is a connected planar garph of order $n \geq 3$ and size $m$ without triangles then $$m \leq 2n-4$$
My attempt:
Since our graph contains no triangles we know that $f \leq 4m$. Now from Eulers identity we get $$n-m+f \leq 4m+n-m$$ $$2 \leq 3m +n$$
I dont see how to procede from here to get the desired result.
On
I assume that the graph is simple (without self loops and parallel edges) with $n$ vertices, $e$ edges and $f$ faces. Since each face is surrounded by atleast 4 edges so, number of edges surrounding $f$ faces is atleast 4f.
Since each edge borders exactly $2$ faces, so number of edges surrounding $f$ faces is exactly 2e.
Then $2e\ge 4f\implies e\ge 2f$.
Use Euler's formula $n-e+f=2$ to get
$e\ge 2(2-n+e)\implies e\le 2n-4$
The inequality $f \le 4m$ is wrong. (Or at least, not the inequality you want to use.)
In general, we relate faces (or vertices) to edges by a counting in two ways argument: If the $i^{\text{th}}$ face has $s_i$ sides, then the sum $s_1 + s_2 + \dots + s_f$ counts each edge twice (once from the inside, once from the outside), so $$\sum_{i=1}^f s_i = 2m.$$ If there are no triangles, then $s_i \ge 4$ for all $i$, so $$2m = \sum_{i=1}^f s_i \ge \sum_{i=1}^f 4 = 4f.$$ So we get the inequality $2m \ge 4f$, or $f \le \frac12 m$.
So $n - m + f = 2$ tells us that $n - m + \frac12m \ge 2$, which you can now solve to get the inequality you wanted.